We integrate by parts, and apply (REF )
\(\begin{split}- \int _M (\Delta _g v, \, i v)_g \, \psi \, \mathrm {vol}_g&= \int _{M} (D v, \, D(\psi \, i v))_g \, \mathrm {vol}_g\\&= \int _{M} (D v, \, iv\otimes \mathrm {d}\psi )_g \, \mathrm {vol}_g+ \int _{M} (D v, \, D(i v))_g \, \psi \, \mathrm {vol}_g\end{split}\) 
Here, \(iv\otimes \mathrm {d}\psi \)  is the (fibre-wise linear) operator
\(TM\rightarrow TM\)  that maps a vector field \(w\)  to the vector field \(i v \, \mathrm {d}\psi (w)\) .
We claim that
\( (D v, \, D(i v))_g = 0 \qquad \textrm {at any point of } M.\) 
Indeed, let \(\left\lbrace \tau _1,\tau _2\right\rbrace \)  be a
moving frame on an open set \(U\subseteq M\) .
There holds that \(\tau _2 = i\tau _1\)  and thus
\(v &= v^1\tau _1 +iv^2\tau _1\\iv &= -v^2\tau _1 +v^1\tau _2.\) 

Therefore (see Section REF  in this paper and [1]}) for \(j=1,2\) ,
\(D_{\tau _j}(iv) &= -D_{\tau _j}(v^2\tau _1) + D_{\tau _j}(v^1\tau _2) \\&=-(\mathrm {d}v^2(\tau _j) -\mathcal {A}(\tau _j)v^2)\tau _1 + (\mathrm {d}v^1(\tau _j) -\mathcal {A}(\tau _j)v^1)\tau _2\\&= iD_{\tau _j} v\,\,\,\,\,\,\hbox{ in }U,\) 

from which (REF ) follows.
Now, we consider \((Dv, \, iv\otimes \mathrm {d}\psi )_g\) .
In terms of the orthonormal frame \(\lbrace \tau _1, \, \tau _2\rbrace \) ,
we have
\((Dv, \, iv\otimes \mathrm {d}\psi )_g= (D_{\tau _1} v, \, i v)_g \, \mathrm {d}\psi (\tau _1)+ (D_{\tau _2} v, \, i v)_g \, \mathrm {d}\psi (\tau _2)\) 
Recalling the definition of \(j(v)\) , Equation REF ,
we can write
\((Dv, \, iv\otimes \mathrm {d}\psi )_g= j(v)(\tau _1) \, \mathrm {d}\psi (\tau _1)+ j(v)(\tau _2) \, \mathrm {d}\psi (\tau _2)= (j(v), \, \mathrm {d}\psi )_g\) 
Therefore,
\(- \int _M (\Delta _g v, \, i v)_g \, \psi \, \mathrm {vol}_g= \int _{M} (j(v), \, \mathrm {d}\psi )_g \, \mathrm {vol}_g\stackrel{(\ref {eq:in_parts_form})}{=}\int _{M} \mathrm {d}^*j(v) \, \psi \, \mathrm {vol}_g\) 
